Question

Repeat Problem 4-8 if a 1.2-V signal is being digitized with the same three ADCs and the input signal is (a) not amplified and (b) amplified by 10 to bring it to full scale.

Short answer

For 8 bits,

$\text{maximumuncertainty}$= 0.04375

For 12 bits,

$\text{maximumuncertainty}=2.734\times {10}^{-3}$

For 16 bits,

$\text{maximumuncertainty}=1.708\times {10}^{-4}$

Step-by-step solution

Part (a)

The analog to digital converters (ADCs) converts analog signals to digital form. The maximum uncertainty can be calculated using following equation.

$\text{maximumuncertainty}=\frac{v_i}{2^n}$

v_i= Input voltage

n= number f bits

For 8 bits,

$\text{maximumuncertainty}=\frac{1.2}{2^8}=4.6875\times {10}^{-3}$

For 12 bits,

$\text{maximumuncertainty}=\frac{1.2}{2^{12}}=2.9297\times {10}^{-4}$

For 16 bits,

$\text{maximumuncertainty}=\frac{1.2}{2^{16}}=1.8310\times {10}^{-5}$

Part (b)

The analog to digital converters (ADCs) converts analog signals to digital form. The maximum uncertainty can be calculated using following equation.

$\text{maximumuncertainty}=\frac{v_i}{2^n}$

v_i= Input voltage

n= number f bits

For 8 bits,

$\text{maximumuncertainty}=\frac{11.2}{2^8}$= 0.04375

For 12 bits,

$\text{maximumuncertainty}=\frac{11.2}{2^{12}}=2.734\times {10}^{-3}$

For 16 bits,

$\text{maximumuncertainty}=\frac{11.2}{2^{16}}=1.708\times {10}^{-4}$